Question 4 of 7: Shape functions of a seven-node transition element, and a Serendipity mesh interface

Question 4: Shape functions of a seven-node transition element, and a Serendipity mesh interface (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A four-cornered parent element with three mid-side nodes added, read directly off the printed figure.

Find. All seven shape functions, a reasoned answer on the inter-element continuity of $N_1$, and the defect in the 2-3-4 interface of the Serendipity mesh.

Approach. Build each mid-side function first as a product that already vanishes at every other node, then correct each bilinear corner function by subtracting its value at each added node times that node's function, and verify the two defining properties.

1. Write the three mid-side functions. A mid-side function must equal one at its own node and zero at all six others. On an edge along $\eta=-1$ with the node at $\xi=0$, the factor $(1-\xi^{2})$ kills both corners of that edge and the factor $(1-\eta)$ kills the whole opposite edge; normalising at the node gives the coefficient $\tfrac12$. Thus $$N_5 = \tfrac{1}{2}\left(1-\xi^{2}\right)(1-\eta),\qquad N_6 = \tfrac{1}{2}(1+\xi)\left(1-\eta^{2}\right),\qquad N_7 = \tfrac{1}{2}\left(1-\xi^{2}\right)(1+\eta)$$ Each is already zero at all four corners and at the other two mid-side nodes, so no correction is needed for these three.
2. Start the corners from the bilinear set. The uncorrected bilinear functions $N_i^{0} = \tfrac{1}{4}\left(1+\xi\xi_i\right)\left(1+\eta\eta_i\right)$ already satisfy $N_i^{0}(\text{node }j)=\delta_{ij}$ at the four corners, but they take non-zero values at the added nodes: $N_1^{0}$ is $\tfrac12$ at node 5; $N_2^{0}$ is $\tfrac12$ at nodes 5 and 6; $N_3^{0}$ is $\tfrac12$ at nodes 6 and 7; and $N_4^{0}$ is $\tfrac12$ at node 7. These are exactly the values that must be removed.
3. Apply the standard corner correction. The correction rule subtracts, from each corner function, its own value at every added node multiplied by that node's shape function: $$N_i = N_i^{0} - \sum_{m=5}^{7} N_i^{0}\!\left(\xi_m,\eta_m\right)N_m$$ This leaves the corner values untouched (every $N_m$ is zero at every corner) while driving each corner function to zero at the added nodes, so the Kronecker-delta property is restored for all seven functions at once.
4. Corner 1 — it touches ONE quadratic edge. Node 1 sits on the quadratic bottom edge and on the linear left edge, so only the node-5 term survives: $$N_1 = \tfrac{1}{4}(1-\xi)(1-\eta) - \tfrac{1}{2}N_5 = \tfrac{1}{4}(1-\eta)\left[(1-\xi) - \left(1-\xi^{2}\right)\right]$$ Factoring $(1-\xi)$ out of the bracket leaves $1-(1+\xi) = -\xi$, so $$\boxed{\;N_1 = -\tfrac{1}{4}\,\xi(1-\xi)(1-\eta) = \tfrac{1}{4}\,\xi(\xi-1)(1-\eta)\;}$$ It has collapsed to a plain one-dimensional quadratic in $\xi$ times a linear factor in $\eta$.
5. Corner 4 — the mirror case. Node 4 sits on the quadratic top edge and the linear left edge, so only the node-7 term survives and the same algebra gives $$\boxed{\;N_4 = -\tfrac{1}{4}\,\xi(1-\xi)(1+\eta)\;}$$ Note that $N_1$ and $N_4$ differ only in the sign inside the $\eta$ factor, as the symmetry of the element about $\eta = 0$ requires.
6. Corners 2 and 3 — they touch TWO quadratic edges. For node 2, both the node-5 and node-6 corrections apply: $$N_2 = \tfrac{1}{4}(1+\xi)(1-\eta) - \tfrac{1}{2}N_5 - \tfrac{1}{2}N_6 = \tfrac{1}{4}(1+\xi)(1-\eta)\bigl[1-(1-\xi)-(1+\eta)\bigr]$$ and likewise for node 3 with the node-6 and node-7 corrections, giving $$\boxed{\;N_2 = \tfrac{1}{4}(1+\xi)(1-\eta)(\xi-\eta-1),\qquad N_3 = \tfrac{1}{4}(1+\xi)(1+\eta)(\xi+\eta-1)\;}$$ These are precisely the standard eight-node Serendipity corner forms — the general rule is that a corner touching two enriched edges keeps the full Serendipity bracket, while a corner touching only one collapses to a single quadratic factor.

7. Verify the set. Evaluating all seven functions at all seven nodes reproduces the identity matrix, $N_i(\text{node }j) = \delta_{ij}$, and summing them gives $\sum_{i=1}^{7}N_i = 1$ for every $(\xi,\eta)$ in the element, which guarantees that a rigid-body translation is represented exactly. The stronger test, $\sum N_i\xi_i = \xi$ and $\sum N_i\eta_i = \eta$, also holds, so the set reproduces any linear field — the completeness requirement for convergence. At the centroid the values are $N_1 = N_4 = 0$, $N_2 = N_3 = -\tfrac14$ and $N_5 = N_6 = N_7 = \tfrac12$: a corner shape function of a transition element is legitimately negative inside the element, which is normal for quadratic sets and not an error.
8. Part (b) — restrict $N_1$ to the two sides through node 1. Node 1 lies on the bottom side 1-5-2 ($\eta=-1$) and on the left side 4-1 ($\xi=-1$). Substituting each in turn, $$N_1\big|_{\eta=-1} = -\tfrac{1}{4}\xi(1-\xi)(2) = \tfrac{1}{2}\xi(\xi-1),\qquad N_1\big|_{\xi=-1} = -\tfrac{1}{4}(-1)(2)(1-\eta) = \tfrac{1}{2}(1-\eta)$$
9. Interpret the restrictions — the answer is yes. On the bottom side, $N_1$ reduces to $\tfrac12\xi(\xi-1)$, which is exactly the one-dimensional quadratic Lagrange function of a three-node line with nodes at $\xi = -1, 0, +1$; the functions of nodes 3, 4, 6 and 7 all vanish identically on that side, so the field there is controlled only by the three nodes 1, 5 and 2 that a neighbouring element shares. On the left side, $N_1$ reduces to $\tfrac12(1-\eta)$, the two-node linear Lagrange function, and the functions of nodes 2, 3, 5, 6 and 7 all vanish, so that side is controlled only by nodes 4 and 1. In both cases the trace of $N_1$ on the side depends solely on nodes lying on that side, and it is the unique polynomial of the order that side supports. A neighbouring element sharing those nodes therefore produces an identical trace, the two fields match everywhere along the shared boundary, and $\boxed{\;N_1\ \text{satisfies }C^{0}\ \text{continuity on both sides through node 1}\;}$
10. Part (c) — diagnose the 2-3-4 interface. Element a is offset by half an element from b and c, so the segment 2–3–4 plays two incompatible roles: for element a it is a single complete edge, with nodes 2 and 4 as its corners and node 3 as its mid-side node, interpolated by one quadratic across the whole segment; for elements b and c it is two separate half-edges, with node 3 as their shared corner and nodes 2 and 4 as their respective mid-side nodes, so the same segment is interpolated by two different quadratics meeting at node 3.

The two descriptions agree at the three shared nodes but nowhere else along the segment, so the displacement (or temperature) field has a gap or overlap between the nodes: the mesh is incompatible or non-conforming, $C^{0}$ continuity is lost across the interface, the element assembly no longer represents a single continuous field, and the mesh fails the patch test — so convergence to the correct solution is no longer guaranteed and the computed stresses near the interface are unreliable. The two accepted cures are to renumber or remesh so that corners meet corners and mid-sides meet mid-sides, or to insert a transition element of exactly the kind derived in part (a) to grade between the two refinement levels while keeping the shared edges compatible. Multi-point constraints tying the free nodes to the neighbouring edge interpolation are a third, software-dependent option.